AJM:6/12/09 Page 1 of 1 Midterm Solutions
Physics 131 Final Solutions Spring 2009
1.
!
F
3=m!
a!
!
F
1!
!
F
2
=15 N !cos 30°ˆ
x!sin 30°ˆ
y
( )
!24 N cos 60°ˆ
x+sin 60°ˆ
y
( )
!40 N sin 20°ˆ
x!cos 20°ˆ
y
( )
=!38.7 N!ˆ
x+9.3 N!ˆ
y
=39.8 N, 166.5°
this direction is 43.5 ° CW from the acceleration vector.
2. Solve the simple one-dimensional, constant accelera tion problem to find that the acceleration is
a=2h
t2=2.5 m/s2
.
Then apply Newton’s second law to both blocks to f ind two equations involving the acceleration and the tension.
Eliminate the tension and solve for
mtable
to find that
mtable =mhanging g/a!1
( )
=6.0 kg
.
3. a) Apply conservation of momentu m to find
Vx=80 g
( )
20 m/s
( )
!60 g
( )
20 m/s
( )
80 g =5.0 m/s
Vy=
!60 g
( )
20 m/s
( )
80 g =!15.0 m/s
Thus
!
V=15.8 m/s,!-71.6°
. The direction is 71.6° CW from th e final velocity of the 60 g rock.
b) Clearly this was not an elastic collision because the 60 g rock has the s ame kinetic energy afterwards and the 80 g
rock has less.
EXTRA CREDIT By constructing the vector diagram for
!!
p
for the 60 g rock, it’s easy to see that
!!
p
is just
2
times the magnitude of its momentum before or af ter the collision. Thus,
F=
!!
p
!t=2 60 g
( )
20 m/s
( )
8.0 ms =212 N
4. Looking at the vertical motion during the first three seconds, show that
vyi =g!t=30 m/s
. Thus, show that
vxi =vi
2!vyi
2=40 m/s
. Thus, looking at the horizontal motion during the entire flight, show that the total time of
flight is
!t=!x
vxi
=8.0 s
. Finally, looking at the vertical mo tion show that
!y="80 m
. Thus, the cliff is 80 m high.
5. Apply Newton’s S econd law to the car at the top of the loop to find that
mv2
r
=.30 N + mg =.80 N
and that, therefore,
the kinetic energy is
1
2
mv2=0.12 J
. Then apply conservation of energy to the period from just before launch to the top
of the loop to find that
1
2kx2=1
2mv2+mg 2R
( )
=.42 J
and that, therefore,
k=2 .72 J
( )
.02 m
( )
2=2100 N/m
.
EXTRA CREDIT Eliminating the 0.30 N contact for ce from step one above, we find
mv2
r
=.50 N
,
1
2
mv2=0.075 J
,
1
2kx2=1
2mv2+mg 2R
( )
=.375 J
, so that
x=2 .375 J
( )
2100 N/m =1.89 cm
6. a) Clearly, the two tensions have to add up to the weight of the beam, 1000 lb .
b) Recognizing that the gravitational force acts at th e center of mass, which will be at the center of the beam and taking
torques around the right end of the beam we find
!
net = +FgL/ 2
( )
"Tleft
4
5L
( )
=0
so
Tleft =5
8
1000 lb
( )
=625 lb
and
Tright =1000 lb !625 lb =375 lb
